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I've seen questions like this on a lot of old qualifying exams in algebraic topology. The type of question is usually of the form:

Let $X$, $Y$ be some topological spaces. Prove that $X$ can't be embedded in $Y$.

The problem I have is that I don't see how homology or homotopy or really anything I know helps. For example, in the general case taking e.g. $S^1$ we can embed it into the disk $D^2$. However, the latter is contractible, so all homotopy and homology groups are zero. Thus, knowing the homotopy/homology groups for the larger space does not give us any information about whether or not we can embed something into it. This would suggest that homotopy and homology doesn't provide any general methods as the spaces in this particular example are pretty much as nice as anything can get.

The question is a somewhat soft question, but I'm curious about which tools in algebraic topology are usually used to tackle questions like this and what are some of the standard tricks involved?

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Hi, I think the material you need is in Hatcher Chapter 2. I am not an expert on this so I cannot suggest further. –  Kerry Mar 11 '11 at 18:53
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Depending on the spaces $X$ and $Y$, you may need very little tools. For example, consider the problem of embedding a circle in $\mathbb R$. One argument that it can't be done is that removal of a point from $\mathbb R$ disconnects it, but no point can disconnect a circle. This is a connectivity argument, and this type of argument persists for all kinds of embedding problems but you may need different types of connectivity results, like homology or homotopy groups, for example. –  Ryan Budney Mar 11 '11 at 18:57
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I know some of those really easy cases. The particular question I had in mind was trying to embed $RP^{2n}$ into $S^{2n+1}$. I've tried looking at cell decompositions in this case... –  Eric Mar 11 '11 at 19:13
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For $\mathbb RP^{2n}$ in $S^{2n+1}$, I suggest considering the relations between the homology of $\mathbb RP^{2n}$ and that of its complement (in a supposed embedding). This gives you a relationship (Alexander duality) between the homology of $\mathbb RP^{2n}$ and that of its complement. –  Ryan Budney Mar 11 '11 at 19:37
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@ Eric: Another way that people often prove non-embedding theorems is by characteristic classes. For example, suppose $\mathbb{R}P^n\subset \mathbb{R}^{n+k}$ is an embedding (or even an immersion). Then for $\tau$ the tangent bundle and $\nu$ the normal bundle, $\tau\oplus \nu$ must be trivial. So their Stiefel-Whitney classes must be inverses in $H^*(\mathbb{R}P^n;\mathbb{Z}/2)$. You can explicitly compute $w(\tau)$, and this tells you what $w(\nu)$ must be, and this gives you a lower bound on $k$. (Cf. Milnor & Stasheff.) –  Aaron Mazel-Gee Mar 24 '11 at 8:50

1 Answer 1

Depending on the spaces $X$ and $Y$, you may need very little tools. For example, consider the problem of embedding a circle in $\mathbb R$. One argument that it can't be done is that removal of a point from $\mathbb R$ disconnects it, but no point can disconnect a circle. This is a connectivity argument, and this type of argument persists for all kinds of embedding problems but you may need different types of connectivity results, like homology or homotopy groups, for example.

For $\mathbb RP^{2n}$ in $S^{2n+1}$, I suggest considering the relations between the homology of $\mathbb RP^{2n}$ and that of its complement (in a supposed embedding). This gives you a relationship (Alexander duality) between the homology of $\mathbb RP^{2n}$ and that of its complement.

In particular check to see how many path-components the complement $C$ has, i.e. $H_0(C)$. Via Alexander duality this has something to do with whether or not $\mathbb RP^{2n}$ is orientable or not. -- Ryan Budney


Another way that people often prove non-embedding theorems is by characteristic classes. For example, suppose $\mathbb{R}P^n\subset \mathbb{R}^{n+k}$ is an embedding (or even an immersion). Then for $\tau$ the tangent bundle and $\nu$ the normal bundle, $\tau\oplus \nu$ must be trivial. So their Stiefel-Whitney classes must be inverses in $H^*(\mathbb{R}P^n;\mathbb{Z}/2)$. You can explicitly compute $w(\tau)$, and this tells you what $w(\nu)$ must be, and this gives you a lower bound on $k$. (Cf. Milnor & Stasheff.)

I guess the point of that, and really also of Ryan's comment about Alexander duality, is to illustrate how even though $\mathbb{R}^{n+k}$ is contractible you can still use topology to prove nonembedding results. Also, note that embedding a closed $n$-manifold into $\mathbb{R}^{n+k}$ is the same as embedding it into $S^{n+k}$ (for $k>0$), so we're really talking about the same thing. -- Aaron Mazel-Gee

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This is a summary of comments, which I think is as much of an answer as this question is going to get. –  2mkgz Dec 30 '14 at 6:53

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